Given an array A[] of size N, find the Maximum Xor Subsequence such that both A [ i ] and A [ N – i – 1 ] belong to this subsequence, where i ranges between [0, N – 1].
Examples:
Input: N = 8, A [ ] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 13
Explanation:
Maximum Xor Subsequence is {1, 3, 4, 5, 6, 8}Input: N = 5, A [ ] = {3, 2, 6, 5, 4}
Output: 6
Explanation:
Maximum Xor Subsequence is {3, 2, 6, 5, 4}
Approach:
Since, both A[i] and A[N-i-1] should be present in the same subsequence, pair them up and then find maximum xor sum. For each valid index i, calculate (A[i] ^ A[N-i-1]) and store it in new array X. Size of this new array will be N/2.
Naive Solution:
The simplest approach after creating array X[] for this problem is to recursively generate all subsequences of X and find the maximum xor subsequence.
Following is the recursive definition of maxXorSubseq(X, N, i):
maxXorSubseq ( X, N, i ) = MAX ( X[ i ] ^ maxXorSubseq ( X, N, i + 1 ), maxXorSubseq (X, N, i + 1) )
The total possible combinations will be 2N.
Time Complexity: O(2N)
Below is the implementation of the above approach:
C++
// C++ implementation// of the above approach#include <bits/stdc++.h>using namespace std;// Returns maximum xorint maxXorSubseq(vector<int> &x, int n, int i){ if(i == n) return 0; return max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, i + 1));}vector<int> compute(vector<int> a, int n){ vector<int> x; // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) x.push_back(a[i] ^ a[n - i - 1]); // If n is odd if(n & 1) x.push_back(a[n / 2]); return x;}// Driver codeint main(){ int n = 8; vector<int> a = { 1, 2, 3, 4, 5, 6, 7, 8 }; // Getting new array x vector<int> x = compute(a, n); int mxXor = maxXorSubseq(x, x.size(), 0); cout << (mxXor); return 0;}// This code is contributed by mohit kumar 29 |
Java
// Java implementation // of the above approach import java.util.*;class GFG{// Returns maximum xorstatic int maxXorSubseq(List<Integer> x, int n, int i){ if(i == n) return 0; return Math.max(x.get(i) ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, i + 1));}static List<Integer> compute(List<Integer> a, int n){ List<Integer> x = new ArrayList<Integer>(); // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) x.add(a.get(i) ^ a.get(n - i - 1)); // If n is odd if((n & 1) == 1) x.add(a.get(n / 2)); return x;}// Driver codepublic static void main(String[] args){ int n = 8; List<Integer> a = Arrays.asList( 1, 2, 3, 4, 5, 6, 7, 8 ); // Getting new array x List<Integer> x = compute(a, n); int mxXor = maxXorSubseq(x, x.size(), 0); System.out.println((mxXor));}}// This code is contributed by offbeat |
Python3
# Python3 implementation# of the above approach# Returns maximum xor def maxXorSubseq(x, n, i): if(i == n): return 0 return max( x[i]^maxXorSubseq( x, n, i + 1), maxXorSubseq( x, n, i + 1))def compute(a, n): x = [] # Calculate a[i]^a[n-i-1] for i in range(n//2): x.append(a[i]^a[n-i-1]) # If n is odd if(n&1): x.append(a[n//2]) return x# Driver codeif __name__ =="__main__": n = 8 a = [1, 2, 3, 4, 5, 6, 7, 8] # Getting new array x x = compute(a, n) mxXor = maxXorSubseq(x, len(x), 0) print(mxXor) |
C#
// C# implementation // of the above approach using System;using System.Collections.Generic;class GFG { // Returns maximum xor static int maxXorSubseq(List<int> x, int n, int i) { if(i == n) return 0; return Math.Max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, i + 1)); } static List<int> compute(List<int> a, int n) { List<int> x = new List<int>(); // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) x.Add(a[i] ^ a[n - i - 1]); // If n is odd if((n & 1) == 1) x.Add(a[n / 2]); return x; } static void Main() { int n = 8; List<int> a = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8 }; // Getting new array x List<int> x = compute(a, n); int mxXor = maxXorSubseq(x, x.Count, 0); Console.WriteLine((mxXor)); }}// This code is contributed by divyeshrabadiya07 |
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Efficient Approach:
Considering the above implementation, the following is a partial recursion tree for input X = [9, 5, 1]
In the above partial recursion tree, maxXorSubseq({1}) is being solved four times. On drawing the complete recursion tree, it can be observed that there are many subproblems that are solved again and again. So this problem has Overlapping Subproblems and recomputation of the same subproblems can be avoided by using Memoization.
For memoization, create an array dp [ ] array and store:
dp[i] = max(x[i] ^ maxXorSubseq(x, n, i + 1), maxXorSubseq(x, n, idx + 1)
This avoids recalculation for previously calculated indices thus optimizing computational complexity.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;vector<int> dp; // Returns maximum xor sumint maxXorSubseq(vector<int> x, int n, int idx){ if (idx == n) { return 0; } // If already precomputed if (dp[idx] != -1) { return dp[idx]; } int ans = 0; ans = max(ans, x[idx] ^ maxXorSubseq(x, n, idx + 1)); ans = max(ans, maxXorSubseq(x, n, idx + 1)); // Store the maximum dp[idx] = ans; return ans;} vector<int> compute(int a[],int n){ vector<int> x; // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) { x.push_back(a[i] ^ a[n - i - 1]); } // If n is odd if ((n & 1) != 0) { x.push_back(a[n / 2]); } return x;}// Driver codeint main(){ int n = 8; int a[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; // Getting new array x vector<int> x = compute(a, n); // Initialize dp array for(int i = 0; i < x.size(); i++) { dp.push_back(-1); } int mxXor = maxXorSubseq(x, x.size(), 0); cout << mxXor << endl; return 0;}// This code is contributed by divyesh072019. |
Java
// Java implementation of the above approachimport java.io.*;import java.util.*;class GFG{ static Vector<Integer> dp = new Vector<Integer>();// Returns maximum xor sumstatic int maxXorSubseq(Vector<Integer> x, int n, int idx){ if (idx == n) { return 0; } // If already precomputed if (dp.get(idx) != -1) { return dp.get(idx); } int ans = 0; ans = Math.max(ans, x.get(idx) ^ maxXorSubseq(x, n, idx + 1)); ans = Math.max(ans, maxXorSubseq(x, n, idx + 1)); // Store the maximum dp.set(idx,ans); return ans;}static Vector<Integer> compute(int[] a,int n){ Vector<Integer> x = new Vector<Integer>(); // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) { x.add(a[i] ^ a[n - i - 1]); } // If n is odd if ((n & 1) != 0) { x.add(a[n / 2]); } return x;}// Driver codepublic static void main(String[] args) { int n = 8; int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 }; // Getting new array x Vector<Integer> x = compute(a, n); // Initialize dp array for(int i = 0; i < x.size(); i++) { dp.add(-1); } int mxXor = maxXorSubseq(x, x.size(), 0); System.out.println(mxXor);}}// This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 implementation# of the above approach# Returns maximum xor sumdef maxXorSubseq(x, n, idx): if(idx == n): return 0 # If already precomputed if(dp[idx]!=-1): return dp[idx] ans = 0 ans = max( ans, x[idx] ^ maxXorSubseq( x, n, idx + 1)) ans = max(ans, maxXorSubseq( x, n, idx + 1)) # Store the maximum dp[idx] = ans return ansdef compute(a, n): x = [] # Calculate a[i]^a[n-i-1] for i in range(n//2): x.append(a[i]^a[n-i-1]) # If n is odd if(n&1): x.append(a[n//2]) return x # Declared dp[] array globallydp = []# Driver codeif __name__ =="__main__": n = 8 a =[1, 2, 3, 4, 5, 6, 7, 8] # Getting new array x x = compute(a, n) # Initialize dp array dp = [-1 for i in range(len(x))] mxXor = maxXorSubseq(x, len(x), 0) print(mxXor) |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class GFG{ static List<int> dp = new List<int>(); // Returns maximum xor sum static int maxXorSubseq(List<int> x, int n, int idx) { if (idx == n) { return 0; } // If already precomputed if(dp[idx] != -1) { return dp[idx]; } int ans = 0; ans = Math.Max(ans, x[idx]^maxXorSubseq(x, n, idx + 1)); ans = Math.Max(ans, maxXorSubseq(x, n, idx + 1)); // Store the maximum dp[idx] = ans; return ans; } static List<int> compute(int[] a,int n) { List<int> x = new List<int>(); // Calculate a[i]^a[n-i-1] for(int i = 0; i < n / 2; i++) { x.Add(a[i] ^ a[n - i - 1]); } // If n is odd if((n & 1) != 0) { x.Add(a[n / 2]); } return x; } // Driver code static public void Main () { int n = 8; int[] a = { 1, 2, 3, 4, 5, 6, 7, 8 }; // Getting new array x List<int> x = compute(a, n); // Initialize dp array for(int i = 0; i < x.Count; i++) { dp.Add(-1); } int mxXor = maxXorSubseq(x, x.Count, 0); Console.WriteLine(mxXor); }}// This code is contributed by rag2127 |
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Time complexity: O(N)
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